Question

Obtain the value of density of earth from the given data:

$$\begin{gathered} g=9.8m{s}^{-2} \\ R=6.38\times {10}^{6}m \\ G=6.67\times {10}^{-11}M.K.S.units. \end{gathered}$$

Answer 1

Step 1: Given data:

$$\begin{gathered} g=9.8m{s}^{-2} \\ R=6.38\times {10}^{6}m \\ G=6.67\times {10}^{-11}M.K.S.units. \end{gathered}$$

Step 2: Writing the formula for Acceleration due to gravity

$$\begin{gathered} g=G\times \frac{M}{R^2} \\ =G\times \frac{\rho \times V}{R^2} \\ =G\times \frac{\rho \times \frac{4}{3}\pi R^3}{R^2} \\ =\frac{4}{3}\times G\times \rho \times \pi \times R \\ \Rightarrow \rho =\frac{3g}{4G\pi R} \\ where, \\ \rho \to density \\ V\to volume \end{gathered}$$

Step 3: Substituting the given values

$$\begin{gathered} \rho =\frac{3g}{4G\pi R} \\ =\frac{3\times 9.8m{s}^{-2}}{4\times 6.67\times {10}^{-11}\times 3.14\times 6.38\times {10}^{6}m} \\ =0.055\times {10}^5 \\ =5500Kg{m}^{-3} \end{gathered}$$

Hence, density